Commuting metric past Dirac spinors?

[auditor]

I'm wondering how in Peskin & Schroeder they go from

$$i\mathcal{M} = {\overline{v}^s^'} (p^{'}) (-ie\gamma^\mu)u^s(p) \left( \frac{-ig_{\mu\nu}}{q^2} \right) \overline{u}^r (k) (-ie\gamma^\nu) v^{r^{'}} (k)$$

at the bottom of page 131 to (5.1) at the top of 132 which reads

$$i\mathcal{M} = \frac{ie^2}{q^2}(\overline{v}(p^{'}) \gamma^\mu u(p) (\overline{u}(k)\gamma_\mu v(k^{'}))$$

Most of the stuff is ok, in particular dropping the spin superscripts. But how does the metric commute with

$$\overline{u}^r (k)$$

? I kind'a remember that the spinors are elements of the SU(2) group and that this might be related to my question. It seems as though the commutator is 0. But if I write out the metric and the Dirac spinor on matrix and vector form respectively I get a matrix product of the form

$$(4 x 4)\cdot (1 x 4)$$

which is undefined. I suspect by doing this I'm mixing apples and oranges, thus my reference to the SU(2) structure. I really don't have time to dwell in group theoretical details right now, although I'm aware that this is the only way to really get QFT. Could anyone please advice? My intuition is that, yes; they do commute, but I want to be sure.

Thanks!

 

[George Jones]

You're thinking too hard.

$g_{\mu\nu}$ is not the metric, it is (for each $\mu$ and $\nu$) a component of the metric, i.e., a number, and, as such, commutes with everything!

 

[auditor]

Wow - that's true. :) Thanks a lot George, really appreciate it!